Question

A train slows down as it rounds a sharp horizontal turn, going from 92.0 km/h to 54.0 km/h in the 16.0 s that it takes to round the bend. The radius of the curve is 130 m. Compute the acceleration at the moment the train speed reaches 54.0 km/h. Assume the train continues to slow down at this time at the same rate.

Answer 1

Calculation of centripetal acceleration(ac):

$a_c=\frac{v^2}{r}$

we can change our speed in terms of m/s

$v=54km/h=54*\frac{1000}{3600} m/s$

now, we can plug these values

$a_c=\frac{(54*\frac{1000}{3600})^2}{130}$

$a_c=1.73077m/s^2$

Calculation of tangential acceleration(at):

$a_t=\frac{\Delta v}{\Delta t}$

now, we can plug values

$a_t=\frac{(54-92)*\frac{1000}{3600} }{16}$

now, we can simplify it

$a_t=-0.65972 m/s^2$

Calculation of resultant acceleration:

$a=\sqrt{(a_c)^2+(a_t)^2}$

now, we can plug values

$a=\sqrt{(1.73077)^2+(-0.65972)^2}$

$a=1.85224m/s^2$.............Answer